§ 6.4NCERT Class 11 · Chemistry · Chapter 6
Solubility Product (Ksp), Precipitation Dynamics & Qualitative Analysis
In a saturated aqueous solution of a sparingly soluble salt, a heterogeneous dynamic equilibrium exists between the un-dissolved crystalline solid and its solvated ions in solution. The equilibrium constant governing this saturation boundary is the solubility product constant (Ksp).
1. Sparingly Soluble Salts and General Formulation of Ksp
Consider a general sparingly soluble salt AxBy dissolving in pure water to yield molar solubility s (mol/L):
AxBy(s) ⇔ x Ay+(aq) + y Bx−(aq)
At equilibrium: [Ay+] = x s and [Bx−] = y s.
Universal Solubility Product Identity:
Ksp = [Ay+]x · [Bx−]y = (x s)x (y s)y = xx · yy · s(x+y)
Ksp = [Ay+]x · [Bx−]y = (x s)x (y s)y = xx · yy · s(x+y)
Formulas for Standard Salt Stoichiometries
| Salt Stoichiometry Type | Representative Examples | Ions Formed | Ksp Expression | Molar Solubility (s) |
|---|---|---|---|---|
| 1:1 Type (AB) | AgCl, BaSO4, CaCO3 | A+ + B− | Ksp = s · s = s2 | s = √Ksp |
| 1:2 or 2:1 Type (AB2 / A2B) | PbCl2, CaF2, Ag2CrO4 | A2+ + 2B− | Ksp = (s)(2s)2 = 4 s3 | s = (Ksp / 4)1/3 |
| 1:3 or 3:1 Type (AB3 / A3B) | Al(OH)3, Fe(OH)3, Ag3PO4 | A3+ + 3B− | Ksp = (s)(3s)3 = 27 s4 | s = (Ksp / 27)1/4 |
| 2:3 Type (A2B3) | As2S3, Ca3(PO4)2 | 2A3+ + 3B2− | Ksp = (2s)2(3s)3 = 108 s5 | s = (Ksp / 108)1/5 |
Ionic Product (Qsp) vs Solubility Product (Ksp) Criterion
Thermodynamic conditions for precipitation, saturation, and dissolution
Qsp < Ksp Qsp = Ksp Qsp > Ksp
[Unsaturated Solution] [Saturated Equilibrium] [Supersaturated State]
No precipitate forms No precipitate forms PRECIPITATION OCCURS!
More solid can dissolve Dynamic equilibrium state Solid separates until Qsp = Ksp
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Common Ion Suppression:
Adding NaCl (common Cl-) to saturated AgCl:
[Cl-] surges → Qsp > Ksp → Equilibrium shifts LEFT ← → AgCl precipitates → Solubility drops!
The Golden Precipitation Rule: A precipitate can form if and only if the ionic product of the participating ions in solution exceeds the solubility product at that temperature: Qsp > Ksp.
2. Common Ion Effect on Solubility
When a soluble salt containing a common ion is added to a saturated solution of a sparingly soluble salt, the equilibrium shifts in the backward direction according to Le Chatelier's principle, dramatically suppressing the solubility of the sparingly soluble salt.
For example, calculate solubility s' of AgCl in 0.1 M NaCl solution (where Ksp = 1.6 × 10−10):
- AgCl(s) ⇔ Ag+(aq) + Cl−(aq) [Ag+] = s', [Cl−] = s' + 0.1 ≈ 0.1 M
- Ksp = [Ag+][Cl−] = s'(0.1) = 1.6 × 10−10
- s' = 1.6 × 10−9 M
In pure water, solubility was s = √(1.6 × 10−10) = 1.26 × 10−5 M. Notice that the presence of 0.1 M common ion Cl− reduced solubility by nearly 10,000 times!
3. Application to Qualitative Inorganic Salt Analysis (Group Separation)
The differential solubility products and controlled common ion suppression form the scientific foundation of qualitative analysis of basic radicals (cations):
| Analytical Group | Cations Present | Group Reagents | Precipitated Form & Underlying Principle |
|---|---|---|---|
| Group I | Ag+, Pb2+, Hg22+ | Dilute HCl | Insoluble Chlorides (AgCl, PbCl2). Low Ksp ensures immediate precipitation. |
| Group II | Cu2+, Pb2+, Hg2+, Bi3+, Cd2+, As3+ | H2S gas in presence of dilute HCl | Sulfides in acidic medium. HCl provides H+ (common ion), suppressing H2S ionization. [S2−] is kept extremely low, sufficient ONLY to exceed the very low Ksp of Group II sulfides, while Group IV sulfides remain dissolved. |
| Group III | Fe3+, Al3+, Cr3+ | NH4OH in presence of solid NH4Cl | Hydroxides (Fe(OH)3, Al(OH)3). NH4Cl provides NH4+ (common ion), suppressing NH4OH ionization. [OH−] is maintained at a low level sufficient ONLY to precipitate Group III hydroxides (Ksp ~ 10−38), leaving Group IV/V hydroxides dissolved. |
| Group IV | Ni2+, Co2+, Mn2+, Zn2+ | H2S gas in presence of NH4OH | Sulfides in alkaline medium. OH− removes H+ from H2S, shifting ionization forward. High [S2−] precipitates Group IV sulfides possessing higher Ksp (~10−15 to 10−22). |
| Group V | Ba2+, Sr2+, Ca2+ | (NH4)2CO3 in presence of NH4OH + NH4Cl | Carbonates (BaCO3, SrCO3, CaCO3). NH4Cl prevents premature precipitation of MgCO3. |
4. Solved Numerical Problems (JEE Main / Advanced / NEET)
Example 1: The solubility product of Ag2CrO4 is 1.1 × 10⁻¹² at 298 K. Calculate its molar solubility in pure water and compare it with the solubility of AgCl (Ksp = 1.8 × 10⁻¹⁰). Which salt is more soluble?
Correct Answer: Solubility of Ag2CrO4 = 6.5 × 10⁻⁵ M; Ag2CrO4 is MORE soluble than AgCl!
Step-by-step Solution:
1. For Ag2CrO4 (A2B type): Ksp = 4 s3.
s = (Ksp / 4)1/3 = [(1.1 × 10−12) / 4]1/3 = (2.75 × 10−13)1/3 = (275 × 10−15)1/3 ≈ 6.5 × 10⁻⁵ M.
2. For AgCl (AB type): Ksp = s2.
s = √(1.8 × 10−10) ≈ 1.34 × 10⁻⁵ M.
3. Comparison: Even though Ag2CrO4 has a smaller Ksp (1.1 × 10−12 vs 1.8 × 10−10), its molar solubility (6.5 × 10−5 M) is nearly 5 times higher than that of AgCl (1.34 × 10−5 M)!
1. For Ag2CrO4 (A2B type): Ksp = 4 s3.
s = (Ksp / 4)1/3 = [(1.1 × 10−12) / 4]1/3 = (2.75 × 10−13)1/3 = (275 × 10−15)1/3 ≈ 6.5 × 10⁻⁵ M.
2. For AgCl (AB type): Ksp = s2.
s = √(1.8 × 10−10) ≈ 1.34 × 10⁻⁵ M.
3. Comparison: Even though Ag2CrO4 has a smaller Ksp (1.1 × 10−12 vs 1.8 × 10−10), its molar solubility (6.5 × 10−5 M) is nearly 5 times higher than that of AgCl (1.34 × 10−5 M)!
Example 2: Equal volumes of 0.002 M CaCl2 and 0.0004 M Na2SO4 solutions are mixed together. Will a precipitate of CaSO4 form? Given Ksp for CaSO4 = 2.4 × 10⁻⁵.
Correct Answer: Qsp = 2.0 × 10⁻⁷ < Ksp; NO precipitate will form.
Step-by-step Solution:
1. When equal volumes are mixed, the total volume doubles, halving the concentration of each ion:
[Ca2+] = 0.002 / 2 = 1.0 × 10−3 M.
[SO42−] = 0.0004 / 2 = 2.0 × 10−4 M.
2. Calculate ionic product Qsp:
Qsp = [Ca2+] [SO42−] = (1.0 × 10−3) × (2.0 × 10−4) = 2.0 × 10⁻⁷.
3. Compare with Ksp:
Qsp (2.0 × 10−7) < Ksp (2.4 × 10−5).
Since Qsp < Ksp, the solution is unsaturated and no precipitation of CaSO4 occurs.
1. When equal volumes are mixed, the total volume doubles, halving the concentration of each ion:
[Ca2+] = 0.002 / 2 = 1.0 × 10−3 M.
[SO42−] = 0.0004 / 2 = 2.0 × 10−4 M.
2. Calculate ionic product Qsp:
Qsp = [Ca2+] [SO42−] = (1.0 × 10−3) × (2.0 × 10−4) = 2.0 × 10⁻⁷.
3. Compare with Ksp:
Qsp (2.0 × 10−7) < Ksp (2.4 × 10−5).
Since Qsp < Ksp, the solution is unsaturated and no precipitation of CaSO4 occurs.
5. Frequently Asked Questions (FAQs)
1. What is the effect of simultaneous solubility of two sparingly soluble salts having a common ion?
When two sparingly soluble salts with a common ion (e.g., AgCl and AgBr) are placed together in water, each suppresses the dissolution of the other via the common ion effect. If x is the solubility of AgCl and y is the solubility of AgBr, then [Ag+] = (x + y). The equations x(x + y) = Ksp(AgCl) and y(x + y) = Ksp(AgBr) must be solved simultaneously.
2. Why does Mg(OH)2 precipitate in Group V if NH4Cl is not added in Group III?
Without NH4Cl, the ionization of NH4OH is not suppressed, producing a high [OH−] that exceeds the Ksp of Mg(OH)2 (~1.8 × 10−11), causing premature precipitation of magnesium in Group III instead of carrying it forward to its correct qualitative group.
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